Answers | Investigation 1
Applications
1. a. Possible answers:
4. Answers will vary. Possible answers:
5. Possible answers: The perimeters are not
the same, because I counted the number
of units around the edge of each figure
and found that their perimeters were
different.
b. The bumper-car ride has an area of
24 m2 , which is the total number of
square meters used to cover the floor
plan of the bumper-car ride. The
perimeter of 22 m is the total number
of rail sections that are needed to
surround the bumper-car ride.
6.
2. Answers will vary. Maximum perimeter
Adding these six tiles reduced the
perimeter of the figure. Only two of the
new tiles have exposed edges, while
together they cover ten previously
exposed edges in the original figure.
for whole-number dimensions is 34 units,
minimum is 16 units. Possible answers:
Perimeter: 22 units
7. P = 4 * 12 ft = 48 ft, A = 12 ft * 12 ft =
144 ft 2
8. P = 22 * 12 ft = 264 ft,
A = 144 ft 2 * 21 = 3,024 ft 2
9. P = 30 * 12 ft = 360 ft,
A = 144 ft 2 * 26 = 3,744 ft 2
Perimeter: 26 units
10. P = 26 * 12 ft = 312 ft,
3. Answers will vary. Maximum area for
A = 144 ft 2 * 20 = 2,880 ft 2
whole-number dimensions is 16 square
units, minimum is 7 square units. Possible
answers:
11. P = 16 units, A = 7 units2
12. P = 16 units, A = 16 units2
13. P ≈11 units, A ≈5.5 units2
14. P = 40 in., A = 100 in.2
Area: 12 square units
15. P = 40 m, A = 75 m2
16. P = 2/ + 2w, A = /
#w
Area: 8 square units
Covering and Surrounding
1
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Investigation 1
Answers | Investigation 1
22. a. Since 40 * 120 = 4,800, the cost of this
17. Check students’ sketches. (See Figure 1.)
model is 4,800 * +95 = +456,000
18. A = 65 cm2 , P = 38 cm
b. 4,800 , 100 = 48 cars
19. A = 36 cm2 , P = 36 cm
23. Designs will vary and costs are dependent
20. a. The next thing that she did was she
on the number of tiles and rail sections
used. Two possible answers:
7 m by 5 m: area = 35 m2 , perimeter =
24 m, cost = +1,650.00
stretched out the string and measured
it. She was finding the perimeter of
the figure.
b. She got the same answer that she got
6 m by 6 m: area = 36 m2 , perimeter =
24 m, cost = +1,680.00
by counting, 18 cm.
c. No, she can’t because the string
Students will have to make a guess to
get started and then alter the guess to
increase or decrease three inter-related
variables. Look for ways that students
proceeded from their first guesses.
method measures length, not area.
Instead, she must count all the squares.
2
21. a. 6 ft * 81
2 ft = 51 ft
b. 29 ft of molding
24. A
c. Two walls have an area of 6 ft * 6 ft =
ft 2 ,
36
and two walls would have an
area of 6 ft * 8.5 ft = 51 ft 2 . The total
surface area would be 36 + 36 + 51 +
51 = 174 ft 2 . You would need 4 pints of
paint because 174 ft 2 , 50 ft 2 = 3.48
and you round up to 4 so that you will
have enough paint.
D
C
B
25. A 4 ft-by-4 ft square requires the least
amount of material for the sides: 16 ft of
board.
d. Check students’ work. Answers will
vary, but students should use processes
similar to those they used in parts (a)–(c).
Students also need to make sure that
they round the number of pints of paint
up to the nearest whole number to
make sure they have enough paint
to cover the walls.
Figure 1
Rectangle
Length (in.)
Width (in.)
Area
(square in.)
Perimeter (in.)
A
5
6
30
22
B
4
13
52
34
C
6 12
8
52
29
Covering and Surrounding
2
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Investigation 1
Answers | Investigation 1
Length (ft)
Width (ft)
1
240
2
120
3
80
4
60
5
48
6
40
8
30
10
24
12
20
15
16
b.
60
50
40
30
20
10
0
x
0 4
8 12 16 20 24 28 32
Length (m)
c. On the table, look for the greatest
(least) number in the perimeter column.
The dimensions will be next to this
entry in the length and width columns.
On the graph, look for the highest
(lowest) point. Then read left to the
perimeter axis to get the perimeter.
The greatest perimeter is 62 meters
(a 1 m * 30 m rectangle). The least
perimeter is 22 meters (a 5 m * 6 m
rectangle).
b. Possible answer: A car needs at least
8 ft for the length, so the 8 ft-by-30 ft
design would probably be too snug.
The 10 ft-by-24 ft, 12 ft-by-20 ft, and
15 ft-by-16 ft designs would all be
appropriate as garages.
27. a.
70 y
Perimeter (m)
26. a.
Length
(m)
Width
(m)
Area
(m2)
Perimeter
(m)
1
30
30
2
15
3
28. a.
62
Length
(m)
Width
(m)
Area
(m2)
Perimeter
(m)
30
34
1
20
20
42
10
30
26
2
10
20
24
5
6
30
22
4
5
20
18
6
5
30
22
5
4
20
18
10
3
30
26
10
2
20
24
15
2
30
34
20
1
20
42
30
1
30
62
Covering and Surrounding
3
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Investigation 1
Answers | Investigation 1
b.
b.
60
130
50
120
40
110
30
100
20
90
10
0
x
0 4
8 12 16 20 24 28 32
Length (m)
c. On the table, look for the greatest
Length
(m)
Width
(m)
Area
(m2)
Perimeter
(m)
1
64
64
130
2
32
64
68
4
16
64
40
8
8
64
32
16
4
64
40
32
2
64
68
64
1
64
130
80
70
60
50
40
(least) number in the perimeter column.
The dimensions will be next to this entry
in the length and width columns. On
the graph, look for the highest (lowest)
point. Then read left to the perimeter
axis to get the perimeter. The greatest
perimeter is 42 meters (a 1 m * 20 m
rectangle). The least perimeter is
18 meters (a 4 m * 5 m rectangle).
29. a.
140 y
Perimeter (m)
Perimeter (m)
70 y
30
20
10
0
x
0
8 16 24 32 40 48 56 64
Length (m)
c. On the table, look for the greatest
(least) number in the perimeter column.
The dimensions will be next to this
entry in the length and width columns.
On the graph, look for the highest
(lowest) point. Then read left to the
perimeter axis to get the perimeter.
The greatest perimeter is 130 meters
(a 1 m * 64 m rectangle). The least
perimeter is 32 meters (a 8 m * 8 m
rectangle).
(See Figure 2.)
Figure 2
Covering and Surrounding
4
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Investigation 1
Answers | Investigation 1
The greatest area is 4 square meters
(a 2 m * 2 m rectangle). The least
area is 3 square meters (a 1 m * 3 m
rectangle).
30. a. 32 m, 14 m
b. It is a 1 m-by-28 m rectangle. It is long
and skinny.
c. It is a 4 m-by-7 m rectangle. It is more
33. a.
compact, or closer to a square.
Length
Width
Area
Perimeter
1
9
9
20
2
8
16
20
3
7
21
20
4
6
24
20
5
5
25
20
6
4
24
20
perimeter of the two rectangles in parts
(b) and (c).
7
3
21
20
8
2
16
20
9
1
9
20
d. The fixed area is 28 m2 . This is the
area of the two rectangles in parts (b)
and (c).
31. a. 24 m2 , 12 m
b. It is a 7 m-by-7 m square.
c. This rectangle is long and skinny; 1 m
by 13 m.
d. The fixed perimeter is 28 m. This is the
32. a.
Length
(m)
Width
(m)
Area
(m2)
Perimeter
(m)
1
3
3
8
2
2
4
8
3
1
3
8
b.
28 y
24
b.
Area (m2)
28 y
Area (m2)
24
20
16
12
8
16
4
12
0
8
x
0
2
4
6
8 10 12 14
Length (m)
4
0
20
x
0
2
4
6
c. On the table, look for the greatest
8 10 12 14
(least) number in the area column. The
dimensions will be next to this entry
in the length and width columns. On
the graph, look for the highest (lowest)
point. Then read down to the length
axis to get the length. Divide the area
by the length to get the width. The
greatest area is 25 square meters (a
5 m * 5 m rectangle). The least area is
9 square meters (a 1 m * 9 m rectangle).
Length (m)
c. On the table, look for the greatest
(least) number in the area column.
The dimensions will be next to this
entry in the length and width columns.
On the graph, look for the highest
(lowest) point. Then read down to the
length axis to get the length. Divide
the area by the length to get the width.
Covering and Surrounding
5
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Investigation 1
Answers | Investigation 1
34. There are no such rectangles. This is
36. a. One possible answer: A 4 unit-by-12
because we need to double the sum of the
length and width. If length and width are
both whole numbers, their sum is a whole
number. Doubling any whole number gives
an even number. Five is odd.
unit rectangle whose perimeter is
32 units.
b. One possible answer: A 4 unit-by-10
unit rectangle whose area is 40 square
units.
35. No; there are always many different
possible perimeters for rectangles with
given areas.
Connections
37. C
47. 31.45
38. Possible answers: A tile on the classroom
48. 49
29
49. 765
32 or 23 32 or 23.906
11
50. 105
or 35
72
24 or 1 24 or 1.4583
51. a. 8
floor is about 1 ft 2 . A coffee table is about
1 yd2 .
39. 1 yd2 is greater. It is 9 ft 2 .
40. They are the same length. 5 * 12 in. =
60 in.
b. 16
41. 12 m is greater because 120 cm is 1.2 m.
c. 6
42. 120 ft is greater because 12 yd is 36 ft.
52. a. Possible answer:
43. They are the same length. 50 cm = 500 mm.
44. Possible answer: One square meter is
greater because a meter is greater than
a yard.
45. The area is the same because she just
shifted one part of the rectangle to
another part of it. The perimeter is longer
because the distance around the new
shape is longer than the original rectangle.
Each brownie is 2 in. by 2 in.
b. Possible answer:
46. a.
b.
Each brownie is 2.5 in. by 2 in.
c. Possible answer:
c.
d. The factors of a number and the
dimensions of the rectangles that can
be made from that number of tiles are
the same. For example, the factors of
25 are 1, 5, and 25, so each number can
be one dimension of a rectangle with
25 square units of area.
Covering and Surrounding
Each brownie is 2 in. by 12
3 in.
6
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Investigation 1
Answers | Investigation 1
60. This is always true, because E + E + E +
d. If they make 20 brownies, then each
could be 2 in. by 21
2 in. The area of the
bottom of the brownie is 5
E = E, O + O + O + O = E, and E + E +
O + O = E, where E stands for an even
number, O for an odd number.
Alternatively, the formula 2(/ + w) shows
that 2 is a factor of any perimeter with
whole-number length and width.
in. 2 .
e. If they make 30 brownies, then each
could be 2 in. by 12
3 in. The area of the
2
bottom of the brownie is 31
3 in. .
53. a. A = 86,250 ft 2 , P = 1,210 ft
2
b. A = 86,250 , 9 = 9,5831
3 yd ,
1
P = 4033 yd
c. 15 ft * 25 ft = 375 ft 2 , 86,250 , 375 =
61. All of them are correct. A rectangle has
four sides: two lengths and two widths.
62. 2 * (5 + 7.5) = 25
63. By using Stella’s formula, 2 * (50 + w) =
196, the width is 48 cm.
230 classrooms
64. Matt is correct because a square has
Note: The shape of the classroom is not
necessarily maintained.
four sides and all the sides (two lengths
and two widths) are equal.
54. a. 38.25 square feet
65. A = 121 in.2 , P = 44 in.
b. Both students are correct. The area
66. A = 156.25 in.2 , P = 50 in.
of any rectangle can be found by
multiplying the length and the width,
regardless of whether the values are
whole numbers or fractions. Nathan
is using the partial products method,
finding the area of the four regions
then finding the sum. He is using
the Distributive Property that was
developed in Prime Time.
67. 2144 = 12 cm
68. a. Possible answer:
cm
Largest rectangle: 5.6
3.5 cm = 1.6
Second-largest rectangle:
3.5 cm
2.15 cm ≈ 1.63
cm
Third-largest rectangle: 2.15
1.3 cm ≈ 1.65
b. Possible answer: The Nautilus shell is
so popular because the dimensions of
its spiral shape are close to the golden
ratio, which makes it visually appealing.
55. The area of a square is side * side and
all sides of a square are equal. Therefore,
the side length is 2169 = 13 and the
perimeter is 4 * 13 = 52.
69. a. Possible answer:
56. No. There are many rectangles that have
cm
Largest rectangle: 7.6
4.7 cm ≈1.617
cm
Second-largest rectangle: 2.5
1.4 cm ≈1.786
an area of 120 cm2 , such as 5 * 24, 2 * 60,
40 * 3, etc.
cm
Third-largest rectangle: 1.1
0.7 cm ≈1.571
Some of the ratios are less than the
golden ratio, and others are greater.
Overall, the ratios are all close in value
to each other and to the golden ratio.
57. F
58. The 36 card tables should be arranged in a
straight line, seating 74 people.
59. a. 1 by 60, 2 by 30, 3 by 20, 4 by 15, 5 by
12, and 6 by 10.
c. About 104 ft; you can approximate
b. 1 by 61
its width by using the golden ratio,
1 : 1.62.
c. 1 by 62 and 2 by 31
d. The factors of a number and the
dimensions of the rectangles that can
be made from that number of tiles are
the same. For example, the factors of
62 are 1, 2, 31, and 62.
Covering and Surrounding
7
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Investigation 1
Answers | Investigation 1
Extensions
70. You may want to ask students to write their
d. 15 tiles can be added. Each figure must
answers as fractions because the patterns
are more obvious.
a. 1
4m
b. Rectangle: side lengths are 1
4 m,
1 m, 1 m; perimeter is 3 m.
4
4
8
c. Rectangle: side lengths are 1
4 m,
1 m, 3 m; perimeter is 7 m.
16
4
8
d. Rectangle: side lengths are 1
4 m,
1 m, 7 m; perimeter is 15 m.
4
16
32
31
e. Perimeter = 32 m
enclose the pentomino in a 4-by-5
rectangle. Possible answers:
1 m,
8
3
16 m,
7
32 m,
74. a.
71. 3 cm * 6 cm rectangle
72. No; there are many different combinations
of lengths and widths that could add up to
a certain perimeter. A square, on the other
hand, has four equal sides, so you can find
the length of one of the sides by dividing
the perimeter by 4.
b. Possible answer: I conducted a
systematic search.
c. This pentomino has the least perimeter,
73. a. Yes. See diagram below.
10 units, because four of the tiles
have two edges joined. All of the
other pentominos have a perimeter
of 12 units.
b. Answers will vary. Possible answers:
75. a. The area of Loon Lake is 38–42 square
units (380,000–420,000 square meters).
The area of Ghost Lake is 34–37 square
units (340,000–370,000 square meters).
c. 3; each of the three tiles must touch
only one edge as they are added.
Possible answers:
b. Answers will vary, but one possibility
is to use a grid such as a 0.5 with
smaller units.
76. a. You could wrap a string around the
lake on the grid and then measure the
string.
b. The perimeter of Loon Lake is 25–26
units (2,500–2,600 m). The area of Ghost
Lake is 45–50 units (4,500–5,000 m).
Covering and Surrounding
8
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Investigation 1
Answers | Investigation 1
77. a. Ghost Lake. Ghost Lake would also
78. a. approximately 31 cm2
make a better Nature Preserve since
it has more shoreline for bird nests, a
variety of vegetation, etc.
b. approximately 40 cm
c. The amount of rubber in the sole is
related to the area of the foot. The
amount of thread required to stitch
the sole to the rest of the shoe is
related to the perimeter (although you
would have to ignore the part of the
perimeter between the toes!).
b. Loon Lake has more room to cruise.
c. Loon Lake is better for swimming,
boating, and fishing.
d. Ghost Lake has more shoreline for
campsites.
Covering and Surrounding
9
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Investigation 1